Error probability of Type I and II

Question

Let $X_1, \ldots , X_{10}$ be independent and $N(\mu, 4)$-distributed.

We consider the Gauss-test with $H_0: \mu \geq 2$ and $H_1:\mu<2$, with the significance level $\alpha=5\%$.

1. Determine the distribution of the test statistic for $\mu=1$ and $\mu=3$.

2. Determine the error probability of type II if $\mu=1$.

3. Determine the error probability of type I if $\mu=3$.

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I have done the following:

1. The test statistic is $$z=\frac{\overline{X}-\mu_0}{\frac{\sigma}{\sqrt{n}}}=\frac{\overline{X}-\mu_0}{\frac{2}{\sqrt{10}}}$$ for $\mu_0=1$ or $\mu_0=3$, or not?

   How is this distributed?

2. The error probability of Type II is equal to the probability that we reject the null hypothesis although it is true. But how could we calculate it?

3. The error probability of Type I is equal to the probability that we accept the null hypothesis although it is wrong. But how could we calculate it?

Answer 1

(1) The distribution of $\overline X$ is $N(\mu, \sigma/\sqrt{n})$. Can you clarify if we have $\sigma=4$ or $\sigma^2=4$?

(2) That is not correct. The type II error $\beta$ is the probability that we keep the null hypothesis even though it is incorrect.

$$\beta = P\left(\frac{\overline X-\mu_0}{\sigma/\sqrt n} > z^* \mid \mu=1\right) =P\left(\frac{\overline X-\mu}{\sigma/\sqrt n} > \frac{\mu_0 -\mu}{\sigma/\sqrt n} + z^* \mid \mu=1\right) =P\left(Z > \frac{2 - 1}{2/\sqrt{10}} - 1.645\right) $$

(3) I'm afraid that is not correct either. The Type I error is simply $\alpha$, which is the probability that we reject the null hypothesis even though it is correct.